Question : If the mean of the following data is 14.7, find the values of p and q.
| Class Intervals (CI) | Frequency (fi) |
| 0-6 | 10 |
| 6-12 |
p |
| 12-18 | 4 |
| 18-24 | 7 |
| 24-30 | q |
| 30-36 | 4 |
| 36-42 | 1 |
| Total | 40 |
Doubt by Gauri
Solution :
| Class Intervals (CI) | Frequency (fi) | Class Marks (xi) | di = xi-a | fidi |
| 0-6 | 10 | 3 | -18 | -180 |
| 6-12 |
p | 9 | -12 | -12p |
| 12-18 | 4 | 15 | -6 | -24 |
| 18-24 | 7 | 21=a | 0 | 0 |
| 24-30 | q | 27 | 6 | 6q |
| 30-36 | 4 | 33 | 12 | 48 |
| 36-42 | 1 | 39 | 18 | 18 |
| Total | Σfi=40 | Σfidi=-138-12p+6q |
Here
Mean (x̄) =14.7 (Given)
Σfi =40
26+p+q=40
p+q=40-26
p+q=14 — (1)
By Assumed Mean Method
x̄ = a + Σfidi/Σfi
14.7 = 21 + (-138-12p+6q)/40
14.7-21 = (-138-12p+6q)/40
-6.3×40=-138-12p+6q
-252=-138-12p+6q
-252+138=-12p+6q
-114=-6(2p-q)
114/6 = 2p-q
19=2p-q
2p-q = 19 — (2)
Solving equation (1) and (2)
p+q=14
2p-q=19
_____________
3p+0=33
_____________
p=33/3
p=11
putting in equation (1)
p+q=14
11+q=14
q=14-11
q=3=
Hence, p=11 & q=3.